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Hoochie [10]
2 years ago
15

Question :

Physics
2 answers:
riadik2000 [5.3K]2 years ago
8 0

The second law of motion states that: the acceleration of an object is dependent upon two variables: - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object

<h3>Meaning of Motion</h3>

Motion can be defined as the process of changing position willingly or due to a force applied.

Motion can be seen in different forms and types depending on the object.

In conclusion, The second law of motion is used to deduce the formula for acceleration.

Learn more about second law of motion: brainly.com/question/2009830

#SPJ1

snow_lady [41]2 years ago
4 0

Answer:

Newton's second law is a quantitative description of the changes that a force can produce on the motion of a body. It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.

Explanation:

Hope it's helpful to you

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This type of light production is an example of _______ energy being transformed into light energy.
Anton [14]

Electric energy??? there isn't much information from the question but I can infer it's electricity.

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3 years ago
The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T
zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than -0.25 m/s^2

f) The two trains avoid collision if the acceleration of the freight train is at least 0.35 m/s^2

Explanation:

a)

We can describe the position of the passenger train at time t with the equation

x_p(t)=u_p t + \frac{1}{2}at^2

where

u_p = 25.0 m/s is the initial velocity of the passenger train

a=-0.100 m/s^2 is the deceleration of the train

On the other hand, the position of the freight train is given by

x_f(t)=x_0 + v_f t

where

x_0=200 m is the initial position of the freight train

v_f = 15.0 m/s is the constant velocity of the train

The collision occurs if the two trains meet, so

x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

b)

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

x_f(t)=x_0 +v_ft

And substituting t = 22.5 s, we find:

x_f(22.5)=200+(15)(22.5)=537.5 m

We can verify that the passenger train is at the same position at the time of the collision:

x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m

So, the two trains collide at x = 537.5 m.

c)

In the graph in attachment, the position-time graph of each train is represented. We have:

  • The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
  • The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

d)

In order to avoid the collision, the freight train must have a initial position of

x_0'

such that the two trains never meet.

We said that the two trains meet if:

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t

Re-arranging,

\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0

Substituting the values for the acceleration and the velocity,

0.05t^2-10t+x_0'=0

The solution of this equation is given by the formula

t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m

Therefore, the freight train must have a head start of 500 m.

e)

In this case, we want to find the acceleration a' of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t

Re-arranging

\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0

Substituting,

-0.5at^2-10t+200=0

The solution to this equation is

t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}

Again, the two trains never meet if the discriminant is negative, so

10^2-4\cdot (-0.5a') \cdot (200)

So, the deceleration must be smaller (towards negative value) than -0.25 m/s^2

f)

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2

where a_f is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2

Re-arranging,

\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0

And the solution is

t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}

Again, the two trains avoid collision if the discriminant is negative, so

10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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3 years ago
If a book has a a mass of 2 kg, how much does the book weigh?
Inessa05 [86]

Answer:

A. 20 N

Explanation:

Weight of the book = mass × acceleration due to gravity

= 2 × 10

= 20 N

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What is observed that connects radio emissions in the galactic nucleus with the emissions in the halo or radio lobes?
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Answer:

What is observed that connects radio emissions in the galactic nucleus with the emissions in the halo or radio lobes? There is a jet of matter coming out of the nucleus, which often points toward the lobes.

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How much heat h1 is transferred to the skin by 25.0 g of steam onto the skin? the latent heat of vaporization for steam is l=2.2
elena-s [515]
The heat transferred by the steam to the skin is given by
Q=m L_v
where
m is the mass of the steam
L_v is the latent heat of vaporization.

In our problem, the mass of the steam is (converting into kg)
m=25.0 g=0.025 kg
while the latent heat of vaporization of the steam is
L_v = 2.256 \cdot 10^6 J/kg
Substituting into the previous formula, we find the heat transferred to the skin:
Q=m L_v = (0.025 kg)(2.256 \cdot 10^6 J/kg)=56400 J = 2.56 \cdot 10^4 J
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